If I have a typical Drude dielectric permittivity:

$$\epsilon(\omega) = \epsilon_{\infty} - \frac{\omega_p^2 \tau}{\omega^2\tau + i\omega}$$

Now, decomposing $\Re{(\epsilon(\omega))}$ and $\Im{(\epsilon(\omega))}$:

$$\Re{(\epsilon(\omega))}=-\frac{\omega_p^2}{\omega^2+1/\tau^2} + \epsilon_\infty$$ $$\Im{(\epsilon(\omega))} = \frac{\omega^2_p}{\omega^3 \tau+1/\tau}$$

Now, the imaginary part is the important one, because it relates the permittivity to conductivity (which is what I want).

$$\epsilon_{\text{im}}(\omega)= \frac{\sigma(\omega)}{\omega \epsilon_0}$$

So, why does equating the imaginary part and the above equation not yield the correct answer?


Should be:

$$\sigma_D(\omega) = \frac{\epsilon_0\omega_p^2 \tau}{1-i\omega\tau}$$

Where am I going wrong?


You’re going wrong by considering the optical conductivity to be real only. Conductivity is complex just as permittivity is! It can be related to epsilon through Ampere’s Law: $$\nabla\times H = J + \frac{dD}{dt}$$ For time-harmonic fields, we get $$\nabla\times H = (\sigma - i \omega \epsilon)E.$$ So the complex relative permittivity is related to the complex conductivity as $$\epsilon_r=\frac{\sigma}{i \omega \epsilon_0} - 1.$$ This is general (although there might be a sign discrepancy depending on your phasor convention) and, depending on your mood, can be taken as a matter of definition. If you leave out the imaginary part of $\sigma$, then you lose half of the information contained in the Drude model!

  • $\begingroup$ So, given the complex relative permittivity and complex conductivity, how can I relate the two such that my last equation appears? $\endgroup$
    – smollma
    Nov 3 '20 at 12:40
  • $\begingroup$ @smollma Try replacing the 1 in my expression with $\epsilon_\infty$ and possibly taking the complex conjugate. Let me know how it goes! $\endgroup$
    – Gilbert
    Nov 3 '20 at 13:03
  • $\begingroup$ are you sure the last expressio nyou have is correct? Please see physics.stackexchange.com/q/591278 your 1 has a sign that would indicate a mismatched time harmonic dependence? $\endgroup$
    – smollma
    Nov 5 '20 at 17:14
  • $\begingroup$ @smollma I’m not sure; feel free to proofread! My larger point is about recognizing the conductivity as complex. Does that make sense? $\endgroup$
    – Gilbert
    Nov 5 '20 at 17:17
  • $\begingroup$ Yes, I understand that the conductivity should be complex. I was trying to equate the imaginary part of the conductivtiy and imag. part of permittivity (ignoring $i$) to use this to swap between the two functions. I can't seem to do it though. So: $$\epsilon_{\infty} +i \frac{\sigma(\omega)}{\omega \epsilon_0} = \epsilon_{\infty} - \frac{\omega_p^2 \tau}{\omega^2 \tau +i\omega}$$ This leaves me with: $$\sigma(\omega) = -i 2 \epsilon_{\infty}\omega\epsilon_0 - \frac{\omega_p^2\omega\epsilon_0\tau}{j(\omega^2\tau +i\omega)}$$ Assume $\epsilon_\infty$ is 1? $\endgroup$
    – smollma
    Nov 5 '20 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.